RMO Part-10!

1)Let A=(a2+4ab+b2)\large(a^{2}+4ab+b^{2}), where a and b are postive integers.

Prove that A≠2015. (HINT-Use Modular Arithmetic or Try to use the concept of Odd-Even numbers)

2)Prove that any prime number, (22n+1)\large(2^{2^n}+1) cannot be represented as a difference of 2 fifth powers of integers.(HINT-Expand the expression and You will get something in common)

3)Find all pairs (x,y) where x and y are integers such that (x3+113=y3)\large(x^{3}+11^{3}=y^{3}).(HINT-Try to apply identities)

4)If (1a+1b+1c=1a+b+c)\large(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c})

Then Prove that,

(1a3+1b3+1c3=1a3+b3+c3)\large(\frac{1}{a^{3}}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^{3}+b^{3}+c^{3}}).(HINT-Factorize )

5)Prove that (22225555+55552222)\large(2222^{5555}+5555^{2222}) is divisible by 7.(HINT-Use Modular Arithmetic or Apply the concept of an±bna^{n}±b^{n}.)

Today I have posted some easy problems ,So I'm sure that you will be able to solve all these 5 problems in 1.5 hours.

Also try my Set RMO.

Note by Naitik Sanghavi
4 years ago

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Here's my solution to (1).

Assume that ,

(a+b)2+2ab=2015\large (a+b)^{2}+2ab=2015

(a+b)2=20152ab\large (a+b)^{2}=2015-2ab

So, (a+b)2\large (a+b)^{2}=Odd

So,(a+b)\large (a+b)=Odd(2q+1)For some positive integer q.

So,2ab=4p\large 2ab=4pFor some integer p.

So, (2q+1)2+4p=2015\large (2q+1)^{2}+4p=2015

4(q2+q+p)=2014\large 4(q^{2}+q+p)=2014

q2+q+p=20144\large q^{2}+q+p=\frac{2014}{4}

Now we can see that R.H.S.is not an integer but In L.H.S. all are intgers .Thus it does not satisfy the given conditions.Hence Proved.

naitik sanghavi - 4 years ago

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1) a2+4ab+b2=0,2or1mod4a^2 + 4ab + b^2 = 0, 2 or 1 mod4 but 2015 = 3 mod4. A contradiction.

Dev Sharma - 4 years ago

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Nice Observation!! I'm having an different solution!!You can try out other problems!!I will post my solutions after some days!!!!!!!

naitik sanghavi - 4 years ago

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Just a small correction, a2+4ab+b2a^2+4ab+b^2 can also be 0(mod4)0 \pmod 4.Try using \pmod for mod. I too did it exactly the same way.

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Yes,of course!! You can post your solutions if you have solved any or have an different solution from others!!!

naitik sanghavi - 4 years ago

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I seems that you have solved all problems on this Set,You can try my other problems of this set and Try my set Olympiad Preparation setfor an easier level!!

naitik sanghavi - 4 years ago

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2) Suppose 22n+1=a5b52^{2^n}+1=a^5-b^5,where a5b5a^5-b^5 be the difference of two fifth powers.a5b5=(ab)(a4+a3b+a2b2+ab3+b4) wherea>ba^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)\ \quad where \quad a>b.

Since 22n+1isprime,ab=12^{2^n}+1 \quad is \quad prime, \quad a-b=1.

By Fermat's Little theorem, 22n+1=a5b5=ab(mod5)=1(mod5)2^{2^n}+1=a^5-b^5=a-b \pmod 5=1 \pmod 5.

Thus 22n+12^{2^n}+1 cannot be represented as a difference of two fifth powers.

Moreover, 22n+12^{2^n}+1 cannot be represented as the difference of two pth powers, where p is an odd prime.

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5) 5555=4(mod7)5555=4 \pmod 7 or 55552222=42222(mod7)5555^{2222}=4^{2222} \pmod 7

43=1(mod7)4^3=1 \pmod 7 or 42220=1(mod7)4^{2220}=1 \pmod 7 or 42222=16=2(mod7)4^{2222}=16=2 \pmod 7

Therefore, 55552222=2(mod7)5555^{2222}=2 \pmod 7.

Similarly,

2222=3(mod7)2222=3 \pmod 7 or 22225555=35555(mod7)2222^{5555}=3^{5555} \pmod 7

33=1(mod7)3^3=-1 \pmod 7 or 35553=(1)1851=1=6(mod7)3^{5553}=(-1)^{1851}=-1=6 \pmod 7 or 35555=54=5(mod7)3^{5555}=54=5 \pmod 7

Therefore, 55552222=5(mod7)5555^{2222}=5 \pmod 7

Therefore 55552222+22225555=2+5=7=0(mod7)5555^{2222}+2222^{5555}=2+5=7=0 \pmod 7

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It seems that you have solved all problems of this note,You can my other problems of this set.Try my set Olympiad Preparation Set for an easier level !!

naitik sanghavi - 4 years ago

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4) let p(x)=x3+lx2+mx+pp(x) = x^3 + lx^2 + mx + p with roots a,b,c then p=mnp = mn and c=bc = -b . Now it easy

Dev Sharma - 4 years ago

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Nice!!You got a different to solve this!

naitik sanghavi - 4 years ago

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Thanks for your contribution for the Brilliant RMO community :)

Swapnil Das - 4 years ago

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Its OK .I am also preparing for RMO so just posting some problems daily..!!!!

naitik sanghavi - 4 years ago

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4) This one is quite long! After a lot of solving I got c=-a or c=-b . Substitute both the values and you get the second equation. Am I missing an easier method?

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Here's my solution to (4). 1a+1b=1a+b+c1c\large \frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}

a+bab=c(a+b+c)(a+b+c)×c\large \frac{a+b}{ab}=\frac{c-(a+b+c)}{(a+b+c)×c}

(a+b)×c×(a+b+c)=(a+b)×ab\large (a+b)×c×(a+b+c)=-(a+b)×ab (Don't remove (a+b) from both sides!Why?See Ahead!!!)

(a+b)×c×(a+b+c)+(a+b)ab=0\large (a+b)×c×(a+b+c)+(a+b)ab=0

(a+b)[c×(a+c)+b×(a+c)]=0\large (a+b)[c×(a+c)+b×(a+c)]=0

(a+b)(b+c)(c+a)=0\large (a+b)(b+c)(c+a)=0

a=-b or b=-c or c=-a.

Now substitute anyone value and it will proved!!

naitik sanghavi - 4 years ago

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@naitik sanghavi Nice!!.I multiplied (ab+bc+ca) with (a+b+c),did a little bit of factoring and got c2+(a+b)c+ab=0c^2+(a+b)c+ab=0 and finally solved the quadratic equation to get the above solution.Your solution is much faster upvoted!!

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You can also generalize it for powers of n and then by induction it will be proved for 3.

Priyanshu Mishra - 4 years ago

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3) Is (-11,0),(0,11) the only solutions?

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Yes!!Correct but you need to prove that there are no other solutions too!!

naitik sanghavi - 4 years ago

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@naitik sanghavi y=x3+1133=(x+11)(x211x+121)3y= \sqrt [3]{x^3+11^3}=\sqrt [3]{(x+11)(x^2-11x+121)} which is only a perfect cube when either of the terms are zero. Hence, getting the above solution.

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Solution to Q3,

Through Fermat's last theorem

az+bz=cza^{ z }+{ b }^{ z }={ c }^{ z }

has no natural number solutions for

z>2z>2

So in,

x3+113=y3{ x }^{ 3 }+{ 11 }^{ 3 }={ y }^{ 3 }

Either one of them is 0 or both of them negative,

if xx and yy are negative,

let x=ax=-a and y=by=-b for positive aa and bb.

So,

b3+113=a3{ b }^{ 3 }+{ 11 }^{ 3 }={ a }^{ 3 }

This has no solutions for positive integers.

So one of xx or yy is 00,

if x=0x=0,y=11y=11 and when y=0y=0,x=11x=-11

Therefore the equation has only two solutions,

(0,11),(11,0)(0,11),(-11,0)

Shauryam Akhoury - 1 year ago

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